JEE Main 2026 — Physics Electromagnetism

For two cells connected in parallel with the same polarity, the equivalent emf $E_{eq}$Eeq​ and equivalent internal resistance $r_{eq}$req​ are given by:

$$E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$Eeq​=r1​+r2​E1​r2​+E2​r1​​

$$r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}$$req​=r1​+r2​r1​r2​​

Substituting the given values ($E_1 = 1$E1​=1 V, $r_1 = 2\,\Omega$r1​=2Ω, $E_2 = 2$E2​=2 V, $r_2 = 1\,\Omega$r2​=1Ω):

$E_{eq} = \dfrac{1 \times 1 + 2 \times 2}{2 + 1} = \dfrac{5}{3}$Eeq​=2+11×1+2×2​=35​ V

$$r_{eq} = \dfrac{2 \times 1}{2 + 1} = \dfrac{2}{3}\,\Omega$$req​=2+12×1​=32​Ω

The current $I$I through the external resistance $R$R is $1$1 A:

$$I = \dfrac{E_{eq}}{R + r_{eq}}$$I=R+req​Eeq​​

$$1 = \dfrac{\dfrac{5}{3}}{R + \dfrac{2}{3}}$$1=R+32​35​​

$$R + \dfrac{2}{3} = \dfrac{5}{3} \Rightarrow R = 1\,\Omega$$R+32​=35​⇒R=1Ω

When the polarity of one cell is reversed, the new equivalent emf $E'_{eq}$Eeq′​ becomes:

$E'_{eq} = \dfrac{|E_1 r_2 - E_2 r_1|}{r_1 + r_2} = \dfrac{|1 \times 1 - 2 \times 2|}{3} = \dfrac{3}{3} = 1$Eeq′​=r1​+r2​∣E1​r2​−E2​r1​∣​=3∣1×1−2×2∣​=33​=1 V

The equivalent internal resistance remains the same, $r_{eq} = \dfrac{2}{3}\,\Omega$req​=32​Ω.

The new current $I'$I′ through the external resistance $R$R is:

$I' = \dfrac{E'_{eq}}{R + r_{eq}} = \dfrac{1}{1 + \dfrac{2}{3}} = \dfrac{1}{\dfrac{5}{3}} = \dfrac{3}{5}$I′=R+req​Eeq′​​=1+32​1​=35​1​=53​ A

Given that the new current is $\dfrac{\alpha}{5}$5α​ A, we have:

$$\dfrac{\alpha}{5} = \dfrac{3}{5} \Rightarrow \alpha = 3$$5α​=53​⇒α=3

Answer: $3$3